If $D_{30}$ is the set of the divisors of 30, $x,\ y \in D_{30}$, we define $x + y = LCM(x,\ y),\ x.y = GCD(x,\ y),x^{'} = \frac{30}{x}$ and $f(x,\ y,\ z) = (x + y).\left( y^{'} + z \right),$ then $f(2,\ 5,\ 15)$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
$D_{30} = \left\{ 1,\ 2,\ 3,\ 5,\ 6,\ 10,\ 15,\ 30 \right\}$
$f(2,\ 5,\ 15) = (2 + 5).\left( 5^{'} + 15 \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 10.\left( \frac{30}{5} + 15 \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because\ \ 2 + 5 = LCM\ of\ (2,\ 5) = 10\ and\ 5^{'} = \frac{30}{5} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 10(6 + 15) = 10.30 = 10$

Q. If $D_{30}$ is the set of the divisors of 30, $x,\ y \in D_{30}$, we define $x + y = LCM(x,\ y),\ x.y = GCD(x,\ y),x^{'} = \frac{30}{x}$ and $f(x,\ y,\ z) = (x + y).\left( y^{'} + z \right),$ then $f(2,\ 5,\ 15)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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